Given vec{A} + vec{B} + vec{C} = vec{0}. The | vedclass

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(D) Given $\vec{A} + \vec{B} + \vec{C} = \vec{0}$. Let $|A| = |B| = x$. Then $|C| = \sqrt{2}x$.From $\vec{A} + \vec{B} = -\vec{C}$,squaring both sides

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$90^\circ, 135^\circ, 135^\circ$

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(D) Given $\vec{A} + \vec{B} + \vec{C} = \vec{0}$. Let $|A| = |B| = x$. Then $|C| = \sqrt{2}x$.From $\vec{A} + \vec{B} = -\vec{C}$,squaring both sides: $A^2 + B^2 + 2AB \cos 	heta_1 = C^2$.Substituting values: $x^2 + x^2 + 2x^2 \cos 	heta_1 = (\sqrt{2}x)^2$.$2x^2 + 2x^2 \cos 	heta_1 = 2x^2 \Rightarrow 2x^2 \cos 	heta_1 = 0 \Rightarrow \cos 	heta_1 = 0 \Rightarrow 	heta_1 = 90^\circ$.Now,consider $\vec{B} + \vec{C} = -\vec{A}$. Squaring both sides: $B^2 + C^2 + 2BC \cos 	heta_2 = A^2$.Substituting values: $x^2 + 2x^2 + 2(x)(\sqrt{2}x) \cos 	heta_2 = x^2$.$2x^2 + 2\sqrt{2}x^2 \cos 	heta_2 = 0 \Rightarrow 2\sqrt{2}x^2 \cos 	heta_2 = -2x^2$.$\cos 	heta_2 = -\frac{1}{\sqrt{2}} \Rightarrow 	heta_2 = 135^\circ$.Since $|A| = |B|$,by symmetry,the angle between $\vec{A}$ and $\vec{C}$ is also $	heta_3 = 135^\circ$.Thus,the angles between the vectors are $90^\circ, 135^\circ, 135^\circ$.

Column $I$	Column $II$
$(A)$ $\vec{A} \cdot \vec{B} = \|\vec{A} \times \vec{B}\|$	$(p)$ $\theta = 45^{\circ}$ or $135^{\circ}$
$(B)$ $\vec{A} \cdot \vec{B} = B^2$	$(q)$ $\theta = 0^{\circ}$
$(C)$ $\|\vec{A} + \vec{B}\| = \|\vec{A} - \vec{B}\|$	$(r)$ $\vec{A} = \vec{B}$
$(D)$ $\|\vec{A} \times \vec{B}\| = AB$	$(s)$ $\theta = 90^{\circ}$

Given $\vec{A} + \vec{B} + \vec{C} = \vec{0}$. The magnitudes of two vectors are equal,and the magnitude of the third vector is $\sqrt{2}$ times that of the other two. What are the angles between the vectors?

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